Question

Solve the equation. $$(x-9)(x-6)=0$$

Short answer

The roots of the equation are \(x=9\) and \(x=6\).

Step-by-step solution

Identify Individual Factors

First, identify the individual factors in the equation. You have two factors in this equation: \(x-9\) and \(x-6\).

Apply the Zero-Product Property

As per the zero-product property, for a product of factors to be zero, at least one of the factors has to be zero. Therefore, set each factor equal to zero and solve for \(x\): \[(x-9)=0 \quad \text{ and } \quad (x-6)=0\]

Solve for Each Factor

Solve each of these equations by adding 9 and 6, respectively to each side of the equations. This gives you two potential solutions for \(x\): \[x=9 \quad \text{ and } \quad x=6\]

Key concepts

Factoring Quadratics

Factoring quadratics is a vital skill in algebra that involves breaking down a quadratic equation into simpler parts, typically into a product of binomials. A quadratic equation is any equation that can be written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \), are constants. To factor such an equation, one essentially looks for two numbers that multiply to give the product \( ac \) and add to give the middle coefficient \( b \).

For instance, consider the equation from the exercise \( (x-9)(x-6)=0 \). This equation is already in a factored form, where the factors are \( x-9 \) and \( x-6 \). These are the binomials obtained by factoring the quadratic equation, and they considerably simplify the solving process. By setting each factor equal to zero, you efficiently use the zero-product property to find the equation's roots.

To improve understanding, let's visualize the factoring with an example. If we had \( x^2 - 15x + 54 \), we would look for two numbers that multiply to 54 and add up to -15. The numbers -9 and -6 fit the criteria, so our factored form is \( (x-9)(x-6) \) which aligns with our exercise.

Solving Quadratic Equations

Solving quadratic equations is a fundamental aspect of algebra. These equations typically involve an \( x^2 \) term and take the general form \( ax^2 + bx + c = 0 \). There are several ways to solve quadratic equations, such as factoring, using the quadratic formula, completing the square, or graphically. Factoring, however, is usually the quickest method when it is possible.

Using the exercise as an example, once you have the factored form \( (x-9)(x-6)=0 \), solving the quadratic equation becomes a matter of finding the values of \( x \) that make each factor equal to zero. These \( x \) values are known as the 'roots' or 'solutions' of the equation. When we set \( x-9 \) and \( x-6 \) to zero, we find that \( x \) can be 9 or 6. In practice, whether you graph the equation or factor it, the intersection points of the curve with the \( x \) axis represent the solutions to the equation.

Algebraic Properties

Understanding algebraic properties is essential for manipulating and solving equations effectively. One such property playing a pivotal role in algebra is the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. This property is crucial when solving quadratic equations through factoring.

The exercise provided utilizes the zero-product property directly. After factoring, you're left with \( x-9 \) and \( x-6 \) set to zero. The next logical step is to apply the property by equating each factor to zero and solving for \( x \) to discover the solutions to the equation. Other essential algebraic properties include the distributive property, which allows the multiplication of an element by a sum of elements, and the commutative property, enabling the rearrangement of terms in addition or multiplication without changing the equation's outcome. These properties, combined with the zero-product property, are foundational tools in the algebraist's toolkit.